1. © 2009 Prentice-Hall, Inc. 7 – 1 Decision Science Chapter 3 Linear Programming: Maximization and Minimization

2. © 2009 Prentice-Hall, Inc. 7 – 2 Introduction Linear programmingLP Linear programming (LP) is a mathematical modeling technique designed to help managers make the most effective use of limited resources such as Machinery, labor, money, time, warehouse space, raw materials

3. © 2009 Prentice-Hall, Inc. 7 – 3 Requirements of a Linear Programming Problem LP has been applied in many areas over the past 50 years LP problems have common properties: maximizeminimize objective function 1.All problems seek to maximize or minimize some quantity (the objective function) constraints 2.The presence of restrictions or constraints that limit the degree to which we can pursue our objective NOT negative 3.Decision variables are NOT negative

4. © 2009 Prentice-Hall, Inc. 7 – 4 Flair Furniture Company The Flair Furniture Co. produces tables and chairs Each table requires 4 hours of carpentry and 2 hours of painting Each chair requires 3 hours of carpentry and 1 hour of painting There are 240 hours of carpentry time available and 100 hours of painting Each table yields a profit of $70 and each chair a profit of $50

5. © 2009 Prentice-Hall, Inc. 7 – 5 Flair Furniture Company The company wants to determine the best combination of tables and chairs to produce to reach the maximum profit HOURS REQUIRED TO PRODUCE 1 UNIT DEPARTMENT ( T ) TABLES ( C ) CHAIRS AVAILABLE HOURS THIS WEEK Carpentry43240 Painting21100 Profit per unit$70$50

6. © 2009 Prentice-Hall, Inc. 7 – 6 HOURS REQUIRED TO PRODUCE 1 UNIT DEPARTMENT( T ) TABLES( C ) CHAIRS AVAILABLE HOURS THIS WEEK Carpentry43240 Painting21100 Profit per unit$70$50 1- Decision Variables: T = Num. of tables, C = Num. of chairs 2- Objective Function: Maximize Z = $70 T + $50 C 3- Subject to the constraints 4 T + 3 C ≤ 240(hours) 2 T + 1 C ≤ 100 (hours) 4- Non negativity: T > 0, C > 0

7. © 2009 Prentice-Hall, Inc. 7 – 7 Flair Furniture Company Solve the following problem Maximize Z = $70 T + $50 C Subject to 4 T + 3 C ≤ 240(hours) 2 T + 1 C ≤ 100 (hours) T, C ≥ 0

8. © 2009 Prentice-Hall, Inc. 7 – 8 HOURS REQUIRED TO PRODUCE 1 UNIT DEPARTMENT( T ) TABLES( C ) CHAIRS AVAILABLE HOURS THIS WEEK Carpentry43240 Painting21100 Profit per unit$70$50 Imagine you decide to produce 100 tables and 100 chair, would be feasible solution? Imagine you decide to produce 40 tables and 50 chair, would be feasible solution?

9. © 2009 Prentice-Hall, Inc. 7 – 9 Graphical Solution to an LP Problem The easiest way to solve a small LP problems is with the graphical solution approach The graphical method only works when there are just two decision variables When there are more than two variables, a more complex approach is needed as it is not possible to plot the solution on a two- dimensional graph The graphical method provides valuable insight into how other approaches work

10. © 2009 Prentice-Hall, Inc. 7 – 10 Graphical Representation of a Constraint 100 – – 80 – – 60 – – 40 – – 20 – – C |||||||||||| 020406080100 T Number of Chairs Number of Tables This Axis Represents the Constraint T ≥ 0 This Axis Represents the Constraint C ≥ 0 Figure 7.1

11. © 2009 Prentice-Hall, Inc. 7 – 11 Graphical Representation of a Constraint The first step in solving the problem is to identify a set or region of feasible solutions To do this we plot each constraint equation on a graph We start by graphing the equality portion of the constraint equations 4 T + 3 C = 240 We solve for the axis intercepts and draw the line

12. © 2009 Prentice-Hall, Inc. 7 – 12 Graphical Representation of a Constraint When Flair produces no tables, the carpentry constraint is 4(0) + 3 C = 240 3 C = 240 C = 80 Similarly for no chairs 4 T + 3(0) = 240 4 T = 240 T = 60 This line is shown on the following graph

13. © 2009 Prentice-Hall, Inc. 7 – 13 Graphical Representation of a Constraint 100 – – 80 – – 60 – – 40 – – 20 – – C |||||||||||| 020406080100 T Number of Chairs Number of Tables ( T = 0, C = 80) ( T = 60, C = 0) Graph of carpentry constraint equation 4 T + 3 C ≤ 240

14. © 2009 Prentice-Hall, Inc. 7 – 14 Graphical Representation of a Constraint 100 – – 80 – – 60 – – 40 – – 20 – – C |||||||||||| 020406080100 T Number of Chairs Number of Tables Any point on or below the constraint plot will not violate the restriction Any point above the plot will violate the restriction (30, 40) (30, 20) (70, 40)

15. © 2009 Prentice-Hall, Inc. 7 – 15 Graphical Representation of a Constraint The point (30, 40) lies on the plot and exactly satisfies the constraint 4(30) + 3(40) = 240 The point (30, 20) lies below the plot and satisfies the constraint 4(30) + 3(20) = 180 The point (30, 40) lies above the plot and does not satisfy the constraint 4(70) + 3(40) = 400

16. © 2009 Prentice-Hall, Inc. 7 – 16 Graphical Representation of a Constraint 100 – – 80 – – 60 – – 40 – – 20 – – C |||||||||||| 020406080100 T Number of Chairs Number of Tables ( T = 0, C = 100) Figure 7.4 ( T = 50, C = 0) Graph of painting constraint equation 2 T + 1 C ≤ 100

17. © 2009 Prentice-Hall, Inc. 7 – 17 Graphical Representation of a Constraint 100 – – 80 – – 60 – – 40 – – 20 – – C |||||||||||| 020406080100 T Number of Chairs Number of Tables Feasible solution region for Flair Furniture Painting Constraint Carpentry Constraint Feasible Region

18. © 2009 Prentice-Hall, Inc. 7 – 18 Graphical Representation of a Constraint For the point (30, 20) in the Feasible area Carpentry constraint 4 T + 3 C ≤ 240 hours available (4)(30) + (3)(20) = 180 hours used Painting constraint 2 T + 1 C ≤ 100 hours available (2)(30) + (1)(20) = 80 hours used For the point (70, 40) in the Infeasible area Carpentry constraint 4 T + 3 C ≤ 240 hours available (4)(70) + (3)(40) = 400 hours used Painting constraint 2 T + 1 C ≤ 100 hours available (2)(70) + (1)(40) = 180 hours used  

19. © 2009 Prentice-Hall, Inc. 7 – 19 Graphical Representation of a Constraint For the point (50, 5) in the Border line Carpentry constraint 4 T + 3 C ≤ 240 hours available (4)(50) + (3)(5) = 215 hours used Painting constraint 2 T + 1 C ≤ 100 hours available (2)(50) + (1)(5) = 105 hours used 

20. © 2009 Prentice-Hall, Inc. 7 – 20 Graphical Representation of a Constraint To produce tables and chairs, both departments must be used simultaneously We need to find a solution that satisfies both constraints simultaneously A new graph shows both constraint plots feasible regionarea of feasible solutions The feasible region (or area of feasible solutions) is where all constraints are satisfied feasible Any point inside this region is a feasible solution infeasible Any point outside the region is an infeasible solution

21. © 2009 Prentice-Hall, Inc. 7 – 21 Isoprofit Line Solution Method We need to find the optimal solution from the possible solutions inside the feasible region. the isoprofit line method Plotting objective (profit) function line for a randomly selected small profit value (e.g., $2,100) Graph the objective function ($2,100 = 70 T + 50 C) (C = 2100/50 = 42 T = 2100/72 = 30) Move the objective function line in the direction of increasing profit while maintaining the slope (parallel) The last point touches the isoprofit line in the feasible region is the optimal solution

22. © 2009 Prentice-Hall, Inc. 7 – 22 100 – – 80 – – 60 – – 40 – – 20 – – C |||||||||||| 020406080100 T Number of Chairs Number of Tables Isoprofit line at $2,100 $2,100 = $70 T + $50 C (30, 0) (0, 42) Isoprofit Line Solution Method

23. © 2009 Prentice-Hall, Inc. 7 – 23 100 – – 80 – – 60 – – 40 – – 20 – – C |||||||||||| 020406080100 T Number of Chairs Number of Tables Four isoprofit lines $2,100 = $70 T + $50 C Last point touch the isoprofit lines Isoprofit Line Solution Method

24. © 2009 Prentice-Hall, Inc. 7 – 24 100 – – 80 – – 60 – – 40 – – 20 – – C |||||||||||| 020406080100 T Number of Chairs Number of Tables Optimal solution to the Flair Furniture problem Optimal Solution Point ( T = 30, C = 40) Maximum Profit Line Isoprofit Line Solution Method

25. © 2009 Prentice-Hall, Inc. 7 – 25 HOURS REQUIRED TO PRODUCE 1 UNIT DEPARTMENT( T ) TABLES( C ) CHAIRS AVAILABLE HOURS THIS WEEK Carpentry43240 Painting and varnishing21100 Profit per unit$70$50 1- Decision Variables: T = 30, C = 40 2- Maximum profit $70 X 30 + $50 X 40 = 4100 3- Subject to the constraints 4 X 30 + 3 X 40 ≤ 240(hours) 2 X 30 + 1 X 40 ≤ 100 (hours) 4- Non negativity: T > 0, C > 0

26. © 2009 Prentice-Hall, Inc. 7 – 26 corner point method A second approach to solving LP problems employs the corner point method It involves looking at the profit at every corner point of the feasible region corner pointsextreme point The mathematical theory behind LP is that the optimal solution must lie at one of the corner points, or extreme point, in the feasible region For Flair Furniture, the feasible region is a four-sided polygon with four corner points labeled 1, 2, 3, and 4 on the graph Corner Point Solution Method

27. © 2009 Prentice-Hall, Inc. 7 – 27 100 – – 80 – – 60 – – 40 – – 20 – – C |||||||||||| 020406080100 T Number of Chairs Number of Tables Figure 7.9 Four corner points of the feasible region 1 2 3 4 Corner Point Solution Method

28. © 2009 Prentice-Hall, Inc. 7 – 28 100 – – 80 – – 60 – – 40 – – 20 – – C |||||||||||| 020406080100 T Number of Chairs Number of Tables 1 2 3 4 3 1 2 4 Point : ( T = 0, C = 0)Profit = $70(0) + $50(0) = $0 Point : ( T = 0, C = 80)Profit = $70(0) + $50(80) = $4,000 Point : ( T = 50, C = 0)Profit = $70(50) + $50(0) = $3,500 Point : ( T = 30, C = 40)Profit = $70(30) + $50(40) = $4,100 Because Point returns the highest profit, this is the optimal solution 3 Corner Point Solution Method

29. © 2009 Prentice-Hall, Inc. 7 – 29 Corner Point Solution Method simultaneous equations method To find the coordinates for Point accurately we have to solve for the intersection of the two constraint lines using the simultaneous equations method, we multiply the painting equation by –2 and add it to the carpentry equation 3 4 T + 3 C =240(carpentry line) – 4 T – 2 C =–200(painting line) C =40 Substituting 40 for C in either of the original equations allows us to determine the value of T 4 T + (3)(40) =240(carpentry line) 4 T + 120 =240 T =30

30. © 2009 Prentice-Hall, Inc. 7 – 30 HOURS REQUIRED TO PRODUCE 1 UNIT DEPARTMENT( T ) TABLES( C ) CHAIRS AVAILABLE HOURS THIS WEEK Carpentry43240 Painting and varnishing21100 Profit per unit$70$50 1- Decision Variables: T = 30, C = 40 2- Maximum profit $70 X 30 + $50 X 40 = 4100 3- Subject to the constraints 4 X 30 + 3 X 40 ≤ 240(hours) 2 X 30 + 1 X 40 ≤ 100 (hours) 4- Non negativity: T > 0, C > 0

31. © 2009 Prentice-Hall, Inc. 7 – 31 Summary of Graphical Solution Methods ISOPROFIT METHOD 1.Graph all constraints and find the feasible region. 2.Select a specific profit (or cost) line and graph it to find the slope. 3.Move the objective function line in the direction of increasing profit (or decreasing cost) while maintaining the slope. The last point it touches in the feasible region is the optimal solution. 4.Find the values of the decision variables at this last point and compute the profit (or cost). CORNER POINT METHOD 1.Graph all constraints and find the feasible region. 2.Find the corner points of the feasible reason. 3.Compute the profit (or cost) at each of the feasible corner points. 4. Select the corner point with the best value of the objective function found in Step 3. This is the optimal solution. 5. Solve for the intersection of the two constraint lines using the simultaneous equations method

32. © 2009 Prentice-Hall, Inc. 7 – 32 Solving Minimization Problems Many LP problems involve minimizing an objective such as cost instead of maximizing a profit function Minimization problems can be solved graphically by first setting up the feasible solution region and then using either the corner point method or an isocost line approach (which is analogous to the isoprofit approach in maximization problems) to find the values of the decision variables (e.g., X 1 and X 2 ) that yield the minimum cost

33. © 2009 Prentice-Hall, Inc. 7 – 33 The Fish Pool Co. is considering buying two different brands of fish feed and blending them to provide a good, low-cost diet for its fishes Minimize cost (in cents) = 2 X 1 + 3 X 2 Subject to: 5 X 1 + 10 X 2 ≥ 90 grams(monthly ingredient A) 4 X 1 + 3 X 2 ≥ 48 grams (monthly ingredient B) 0.5 X 1 ≥ 1.5 grams (monthly ingredient C) X 1, X 2 ≥ (non negativity constraint) Fish Pool Co X 1 = number of Kg. of brand 1 purchased to feed fish X 2 = number of Kg. of brand 2 purchased to feed fish Let

34. © 2009 Prentice-Hall, Inc. 7 – 34 Fish Pool Co INGREDIENT COMPOSITION OF EACH POUND OF FEED (OZ.) Minimum monthly requirements PER TURKEY (Gram) BRAND 1 FEEDBRAND 2 FEED A51090 Kg. B4348 Kg. C0.50 1.5 Kg. Cost per Kg. $ 2$ 3 To draw the graph, plot the constraint line 5 X 1 + 10 X 2 ≥ 90 grams X 1 = 18, X 2 = 9 4 X 1 + 3 X 2 ≥ 48 grams X 1 = 12, X 2 = 16 0.5 X 1 ≥ 1.5 grams X 1 = 3

35. © 2009 Prentice-Hall, Inc. 7 – 35 Using the corner point method First we construct the feasible solution region The optimal solution will lie at one of the corners Fish Pool Co. – 20 – 15 – 10 – 5 – 0 – X2X2 |||||| 510152025 X1X1 Kilos of Brand 2 Kilos of Brand 1 Ingredient C Constraint X 1 = 3 Ingredient B Constraint X 1 = 12, X 2 = 16 Ingredient A Constraint X 1 = 18, X 2 = 9 Feasible Region a b c

36. © 2009 Prentice-Hall, Inc. 7 – 36 Using the corner point method First we construct the feasible solution region The optimal solution will lie at one of the corners Fish Pool Co. – 20 – 15 – 10 – 5 – 0 – X2X2 |||||| 510152025 X1X1 Kilos of Brand 2 Kilos of Brand 1 Ingredient C Constraint X 1 = 3 Ingredient B Constraint X 1 = 12, X 2 = 16 Ingredient A Constraint X 1 = 18, X 2 = 9 Feasible Region a b c

37. © 2009 Prentice-Hall, Inc. 7 – 37 Fish Pool Co. We solve for the values of the three corner points Point a is the intersection of ingredient constraints C and B 4 X 1 + 3 X 2 = 48 X 1 = 3 Substituting 3 in the first equation, we find X 2 = 12 + 3 X 2 = 48, then X 2 = (48-12)/3 = 12 Substituting these value back into the objective function we find Cost= 2 X 1 + 3 X 2 Cost at point a = 2(3) + 3(12) = 42

38. © 2009 Prentice-Hall, Inc. 7 – 38 Fish Pool Co. Point b is the intersection of ingredient constraints A and B B 4 X 1 + 3 X 2 = 48 A 5 X 1 + 10 X 2 = 90 Solving for point b with basic algebra Multiply B x 10 40 X 1 + 30 X 2 = 480 Multiply A x -3 -15 X 1 - 30 X 2 = -270 Add the equations 25 X 1 = 210 we find X 1 = 8.4 and X 2 = 4x8.4 + 3 X 2 = 48 X 2 =(48-33.6)/3 = 4.8 Substituting these value into the objective function Cost= 2 X 1 + 3 X 2 Cost at point b = 2(8.4) + 3(4.8) = 31.2

39. © 2009 Prentice-Hall, Inc. 7 – 39 Fish Pool Co. Point c is the intersection of ingredient constraint A when X 2 = 0 A 5 X 1 + 10 X 2 = 90 when X 2 = 0 X 1 = 18 Substituting these value back into the objective function we find Cost= 2 X 1 + 3 X 2 Cost at point c = 2(18) + 3(0) = 36

40. © 2009 Prentice-Hall, Inc. 7 – 40 Cost= 2 X 1 + 3 X 2 Cost at point a= 2(3) + 3(12) = 42 Cost at point b= 2(8.4) + 3(4.8) = 31.2 Cost at point c= 2(18) + 3(0) = 36 Fish Pool Co. The lowest cost solution is to purchase 8.4 Kg. of brand 1 feed and 4.8 Kg. of brand 2 feed for a total cost of $ 31.2

41. © 2009 Prentice-Hall, Inc. 7 – 41 Using the isocost approach Choosing an initial cost of $ 54 it is clear improvement is possible Fish Pool Co. – 20 – 15 – 10 – 5 – 0 – X2X2 |||||| 510152025 X1X1 Pounds of Brand 2 Pounds of Brand 1 Feasible Region 54 = 2 X 1 + 3 X 2 Isocost Line Direction of Decreasing Cost 31.2 = 2 X 1 + 3 X 2 ( X 1 = 8.4, X 2 = 4.8)